Ben Wolfe
11/3/14
Condenser Analysis – Water Cooled
Model:
For this condenser design there will be a coil of stainless steel tubing suspended in a bath of cold water.
The cold water will be stationary and begin at an ambient temperature of 25C. As the steam flows
through the condensing coil, the heat will be transferred from the steam into the surrounding cold
water bath. The cold water bath is assumed to have adiabatic boundary conditions, so all of the heat
loss form the condensing pipe will remain in the water bath, causing the water to gradually heat up over
the 9 hour cycle.
Assumptions:
 The water bath will have an adiabatic boundary condition meaning it will not lose any heat to
the environment.
 Radiation will be neglected.
 Steam flows into the condenser at a constant rate.
 Steam enters the condenser at 200C.
 The temperature of the outside of the pipe wall will be isothermal and equal to the temperature
of the water bath.
 The water bath will heat up uniformly, meaning the bulk temperature is constant throughout
the entire water bath.
 The temperature of the inside surface of the condensing pipe is equal to the temperature of the
water bath.
Approach:
The approach being used to solve this heat transfer problem is to break down the condenser into three
different sections that all require a different set of heat transfer equations.
Step 1: Steam enters the condenser at 200C and must be cooled to 100C
Step 2: The steam remains at 100C as enough heat is removed to condense steam into liquid water.
Step 3: The liquid water cools below 100C and exits the condenser.
The length of pipe required to complete each step will be calculated. The goal of this condenser design is
to minimize the length of pipe required, because that will lower the total cost of the system.
Ben Wolfe
11/3/14
Figure 1: Theoretical diagram showing the each step of heat loss.
Outer Diameter
Wall thickness
Inner Diameter
Pipe Length
Surface Area inner pipe
Temperature of cold water bath
Initial Temperature of steam
Volumetric Flow rate (from boiler-liquid)
Volumetric Flow rate (from boiler-steam)
Mass flow rate (m dot)
velocity of steam (in condenser tube)
velocity of water (after condensation)
0.125
0.02
0.085
240
inches
inches
inches
inches
25
200
1.08
Celsius
Celsius
gal/9 hours
Density of water
Density of steam
viscosity of steam
thermal conductivity (steam)
Prandtl number (steam)
Cp (Specific heat of steam)
viscosity of water (at 75C)
thermal conductivity (water)
Cp (Specific heat of water)
Hv (heat of vaporization of water)
Table 1: Material Properties
1000
0.598
1.23E-05
0.02508
1
1970
0.000378
0.67
4186
2260000
kg/m^3
kg/m^3
kg/m*s
W/m*K
dimensionless
J/kg*K
kg/m*s
W/m*K
J/kg*K
J/kg
0.00318
0.000508
0.00216
6.096
0.0413
298
473
1.2618E-07
0.000211
0.000126
57.64
0.0345
meters
meters
meters
meters
meters^2
Kelvin
Kelvin
m^3/s
m^3/s
kg/s
m/s
m/s
Ben Wolfe
11/3/14
Equations used:
Valid for isothermal wall condition
ReD =
ρ∗v∗D
μ
Step 1 Calculations:
The convection coefficient for internal flow can be calculated using the Nusselt number, which depends
on the Reynolds number. The Nusselt number increases with the Reynolds number, so a high Reynolds
number is more desirable. To achieve a high Reynolds number, the velocity of the moving fluid must be
increased. The fluid velocity increases with a decreasing pipe diameter, so a small pipe diameter is
chosen.
Re
Nu
h bar (convection coefficient)
6059.70
24.42
283.64
turbulent
turbulent
W/m^2*K
Find: how much pipe length it takes for the steam to reach 100°C
L1 ∗ h̅ ∗ π ∗ D ∗ ΔTlm = ṁ ∗ Cp ∗ (𝑇0 − 𝑇𝑖 )
Where T0 = 100C and Ti = 200C
Rearranging this equation to solve for L1 yields:
ΔTi
CP
L1 = ln (
) ∗ ṁ ∗
ΔTo
h̅ ∗ π ∗ D
L1 = 4.3 inches = .011 meters
Ben Wolfe
11/3/14
Step 2 Calculations:
Find: how much pipe length it takes to condense steam at 100°C
I can find how much power is needed to condense the steam by knowing the mass flow rate and the
heat of vaporization of water. Then by setting this value equal to the convection inside the pipe the
length of pipe required for condensing can be found.
Power needed from heat of vaporization
q2 = ṁ ∗ Hv
q2 = 285.2 W
285.2 = h̅ ∗ As ∗ (100°C − Tc )
285.2 = h̅ ∗ π ∗ D ∗ L2 ∗ (100°C − Tc )
Where Tc is the temperature of the water bath, and h̅ is the convection coefficient from step 1.
L2 = 78 inches = 1.98 meters
Step 3 Calculations:
Finally, the steam is fully condensed and cools in the remainder of the pipe. There is no requirement for
the exiting temperature of the fluid, but it is calculated because heat is still being added to the water
bath. The Reynolds number, Nusselt number, and convection coefficient must be recalculated because
the fluid is no longer steam, it is liquid water.
Re
Nu
h bar (convection coefficient)
pipe length (L3)
As (Surface area inside pipe)
To = Ts - (Ts-Ti)/(e^(h*As/(mdot*Cp))
196.7
3.66
1135.8
157.3
laminar
laminar
W/m^2*K
inches
25
Celsius
3.9957
0.0271
meters
meters^2
A total pipe length of 20 ft is assumed to be submerged in the water bath so L3 is simply the remaining
length of pipe. The pipe used for this condenser can only be purchased in 25ft length increments, so it
would be best to avoid going over 25 feet of required pipe for cost reasons. 20 is being used for this
analysis rather than 25ft, because some of the pipe will be used to connect to the boiling chamber and
will therefore not be submerged in the water bath.
L3 = 20ft – L2 – L1
L3 = 157.3 inches = 4.0 meters
Liquid water should exit the condenser at 25C
Ben Wolfe
11/3/14
Transient Analysis
The previous calculations were only valid when the cold water bath is at 25C, but over time this water
will heat up and a transient analysis is necessary to understand the full functionality of this condenser.
Assuming all of the heat loss from the steam enters the water bath and assuming there are 14 gallons of
available water the following chart shows the results of this transient analysis.
Temp
Step 1
Step 2
Step 3
T out
𝑞1 + 𝑞2
+ 𝑞3
of
Liquid
water
water
bath
Tc
L1 (m) q1 (W) L2 (m) q2 (W) L3 (m) q3 (W) Temp (C) qtotal (W)
0
25.0
0.11 24.86
1.98 285.20
4.01
39.6
25
349.67
0.5
27.8
0.11 24.86
2.05 285.20
3.93
38.8
27.8
348.89
1
30.7
0.12 24.86
2.14 285.20
3.84
38.1
30.7
348.12
1.5
33.5
0.12 24.86
2.23 285.20
3.75
37.3
33.5
347.35
2
36.3
0.12 24.86
2.33 285.20
3.65
36.5
36.3
346.57
2.5
39.1
0.13 24.86
2.43 285.20
3.54
35.7
39.1
345.80
3
41.8
0.13 24.86
2.55 285.20
3.42
35.0
41.8
345.03
3.5
44.6
0.13 24.86
2.68 285.20
3.29
34.2
44.6
344.27
4
47.4
0.14 24.86
2.82 285.20
3.14
33.4
47.4
343.50
4.5
50.1
0.14 24.86
2.97 285.20
2.98
32.7
50.1
342.74
5
52.8
0.15 24.86
3.14 285.20
2.81
31.9
52.8
341.98
5.5
55.6
0.15 24.86
3.34 285.20
2.61
31.2
55.6
341.22
6
58.3
0.16 24.86
3.55 285.20
2.39
30.4
58.3
340.46
6.5
61.0
0.16 24.86
3.80 285.20
2.14
29.6
61.0
339.70
7
63.6
0.17 24.86
4.08 285.20
1.85
28.9
63.6
338.95
7.5
66.3
0.18 24.86
4.40 285.20
1.52
28.1
66.3
338.19
8
69.0
0.19 24.86
4.78 285.20
1.13
27.4
69.0
337.44
8.5
71.6
0.20 24.86
5.22 285.20
0.68
26.6
71.6
336.69
9
74.3
0.20 24.86
5.76 285.20
0.13
25.9
78.0
335.94
Table 2: Transient analysis of with 14 gallons of water and no heat loss to the environment
Time
(hours)
From this data it can be seen that the length of pipe required for step 2 increases substantially as the
temperature of the water bath increases. After 9 hours of continuous operation, the water bath will
reach a temperature of 74.3C and the required pipe length will be approximately 19.5 ft before all of the
steam condenses.
Ben Wolfe
11/3/14
Conclusions:
This condenser is much cheaper because there is less stainless steel tubing required than the air
cooled condenser. As a result it is a more feasible design.
This analysis has been conservative in its approach for several reasons. Heat losses to the
environment were neglected entirely, meaning the water bath should be losing heat constantly and the
temperature shouldn’t reach as high as predicted. Also, radiation was neglected entirely which should
again benefit the condenser in the real world because of these heat losses.
Bill of Materials:
Description
Cost
25ft 1/8" tubing
$43.48
18 gallon tote
$9.87
Compression Fitting -1/8"
epoxy sealing
$15.07
$0.00
sealing grommet 1/8" (pkg
of 50)
$1.74
Hex locknut 1/8" NPT
$3.03
source
http://www.amazon.com/gp/product/B00GPAK0KO/ref=ox_sc_act_tit
le_1?ie=UTF8&psc=1&smid=A192LS2FEVGCGZ
http://www.homedepot.com/p/Rubbermaid-18-Gal-15-9-10-in-x-16-12-in-x-23-9-10-in-Storage-Tote-in-Green-1823619/203297506
http://www.amazon.com/dp/B004M18T7O/ref=biss_dp_t_asn
RIT machine shop
Grainger Item # 3MPL2
http://www.amazon.com/gp/product/B003GSL4IK/ref=ox_sc_act_title
_1?ie=UTF8&psc=1&smid=ATVPDKIKX0DER
Total Cost of condenser = $73.19
Sources:
Properties of Stainless Steel
http://www.aksteel.com/pdf/markets_products/stainless/austenitic/304_304l_data_sheet.pdf
Properties of Steam
http://www.thermopedia.com/content/1150/